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- A certain steel has proportionality limit of 300 N/mm² in simple tension. Under a Three Dimensional Stress System, the principal stresses are 150 N/mm² (Tensile), 75 N/mm² (Tensile) and 30 N/mm² (Compressive), μ = 0.3. The factor of safety according to maximum shear stress theory would be1. A specimen of steel 20 mm diameter with a gauge length of 200 n.m is tested io destruction. It' has an extension of 0.25 mm under a load of 80 kN and the load at elasti limit is 102 kN The maximum load is 130 kN. The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find (i) The stress at elastic limit. (ii) Young's modulus. (iii) Percentage elongation. (iv) Percentage reduction in area. (v) Ultimate tensile stress= Syc = 100 kpsi and a true strain at fracture = 0.55. Estimate the factor of safety for the following principal stress states: A hot-rolled steel has a yield strength of Syf of ɛf = | (b) ox = 60 kpsi, ơy |(c) ơx |(d) ox = -40 kpsi, ơy |(a) 0x = 70 kpsi, ơy = 70 kpsi, txy 40 kpsi, txy 40 kpsi, txy = = -60 kpsi, Txy = 0 kpsi -15 kpsi 45 kpsi 0 kpsi, ơy = %3D 15 kpsi | (e) 01 = 30 kpsi, o2 = 30 kpsi, 03 = 30 kpsi
- A high-strength steel has a yield strength of 1380 MPa and fracture toughness of 91 MPavm. A surface crack of 2.5 mm is found at the surface. At what max. applied stress level will catastrophic failure occur? (Y = 1.00 for internal crack, and Y=1.12 for surface crack)A steel specimen is tested in tension. The specimen is 1.0 in. wide by 0.25 in. thick in the test region. By monitoring the load dial of the testing machine, it was found that the specimen yielded at a load of 12.5 kips and fractured at 17.5 kips. a. Determine the tensile stresses at yield and at fracture. b. Estimate how much increase in length would occur at 60% of the yield stress in a 2-in. gauge length. Step-by-step solution: Step 1 of 4 Given that: Width of the specimen, b = 1 in Thickness of the specimen, t = 0.25 in Yield load on the specimen, Py = 12.5 kips Fracture load on the specimen, Pf = 17.5 kips Gauge length, L = 2 in Percentage of yield stress = 60%1. A specimen of steel 20 mm diameter with a gauge length of 200 n.m is tested io destruction. It' has an extension of 0.25 mm under a load of 80 kN and the load at elastic limit is 102 kN The maximum load is 130 kN. The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find (i) The stress at elastic limit. (ii) Young's modulus. (iii) Percentage elongation. (iv) Percentage reduction in area. (v) Ultimate tensile stress
- A steel specimen is tested in tension. The specimen is 1.0 in. wide by 0.25 in. thick in the test region. By monitoring the load dial of the testing machine, it was found that the specimen yielded at a load of 12.5 kips and fractured at 17.5 kips.a. Determine the tensile stresses at yield and at fracture.b. Estimate how much increase in length would occur at 60% of the yield stress in a 2-in. gauge length.Problem 4: A tensile test is carried out on a bar of a mild steel of diameter 2 cm. the bar yields under a load of 150 kN and breaks finally at a load of 70 kN. Estimate; 1-the tensile stress at the yield point 2-the ultimate tensile stress 3-the average stress at the breaking point, if the diameter of the fractured neck is 1 сm.7) A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of 0.25 mm under a load of 80 kN and the load at elastic limit is 102 kN. The maximum load is 130 kN. The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find (i) The stress at elastic limit. (ii) Young's modulus. (iii) Percentage elongation. (iv) Percentage reduction in area. (v) Ultimate tensile stress.
- 2. The Goodman diagram relates oa and om for fatigue failure after a specific number of cycles Nf, where da is the cyclic stress amplitude, and om the mean stress. For a steel specimen it is found that oa oa (0). [1- (om/OTS)] where Ors is the metal's tensile stress (375MPa), and oa (0)~0.450TS is the 107 cycle fatigue limit at zero mean stress. Assuming the specimen is cycled repeatedly between 0 stress and a peak stress, what is the maximum peak stress if failure in < 107 cycles is to be avoided? Ans: 233 MPaQ3: A cylindrical specimen of steel having an original diameter of (12.8mm) is tensile tested to fracture and found to have engineering fracture strength of (450MP ). If its cross-sectional diameter at fracture is (10.7mm), determine: (1) the ductility in terms of percent reduction in area and (2) the true stress at fracture.2- What is the largest size (mm) internal through crack that a thick plate of aluminium alloy 7075-T651 can support at an applied stress of (a) three-quarters of the yield strength and (b) one-half of the yield strength? Assume Y = 1. for 7075-T651, KỊC = 24.2 MPa ym and oYS = 495 MPa.